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Pani-rosa [81]
3 years ago
5

Can someone help me on this

Chemistry
1 answer:
xz_007 [3.2K]3 years ago
8 0
155,500



I did this to the best of my ability. I have a hard time comprehending things sometimes so I’m so so so sorry if it’s wrong
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Consider mixture C, which will cause the net reaction to proceed in reverse. Concentration (M)initial:change:equilibrium:[XY]0.2
Allushta [10]

This is an incomplete question, here is a complete question.

Calculating equilibrium concentrations when the net reaction proceeds in reverse Consider mixture C, which will cause the net reaction to proceed in reverse.

The chemical reaction is:

XY(g)\rightleftharpoons X(g)+Y(g)

Concentration(M)        [XY]            [X]            [Y]

(M)initial:                     0.200        0.300      0.300

change:                         +x               -x              -x

equilibrium:             0.200+x      0.300-x     0.300-x

The change in concentration, x, is positive for the reactants because they are produced and negative for the products because they are consumed. Based on a Kc value of 0.140 and the data table given, what are the equilibrium concentrations of XY, X, and Y, respectively?

Answer : The Equilibrium concentrations of XY, X, and Y is, 0.104 M, 0.204 M and 0.204 M respectively.

Explanation :

The chemical reaction is:

                                 XY(g)\rightleftharpoons X(g)+Y(g)

initial:                      0.200      0.300   0.300

change:                    +x             -x           -x

equilibrium:      (0.200+x)  (0.300-x)   (0.300-x)

The equilibrium constant expression will be:

K_c=\farc{[X][Y]}{[XY]}

Now put all the given values in this expression, we get:

0.140=\frac{(0.300-x)\times (0.300-x)}{(0.200+x)}

By solving the term 'x', we get:

x = 0.0963 and x = 0.644

We are neglecting the value of x = 0.644 because the equilibrium concentration can not be more than initial concentration.

Thus, the value of x = 0.0963

Equilibrium concentrations of XY = 0.200+x = 0.200+0.0963 = 0.104 M

Equilibrium concentrations of X = 0.300-x = 0.300-0.0963 = 0.204 M

Equilibrium concentrations of X = 0.300-x = 0.300-0.0963 = 0.204 M

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This type of erosion picks up materials and transports them in the air, often causing the abrasion of surfaces.
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Answer: Sediment Transport by Wind

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How has the taxonomy of organisms changed over time
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Answer:

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Explanation:

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Explanation:

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